Lions - Nancy L. #82
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Lions - Nancy L. #82
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alope107
reviewed
Oct 18, 2022
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| available_letters = { | ||
| "A": 9, "B": 2, "C": 2, | ||
| "D" : 4, "E": 12, "F": 2, | ||
| "G": 3, "H": 2, "I": 9, | ||
| "J": 1, "K": 1, "L":4, | ||
| "M": 2, "N": 6, "O": 8, | ||
| "P": 2, "Q": 1, "R": 6, | ||
| "S": 4, "T": 6, "U": 4, | ||
| "V": 2, "W":2, "X": 1, | ||
| "Y": 2, "Z": 1 | ||
| } |
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| for _ in range(10): | ||
| letters = [] | ||
| for letter, quantity in available_letters.items(): | ||
| for _ in range(quantity): | ||
| letters.append(letter) | ||
|
|
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| selection = random.choice(letters) | ||
| hand_available.append(selection) | ||
| available_letters[selection] = available_letters[selection] - 1 |
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This works well! Nice use of a list of all the available letters to make random.choice work.
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| # iterate through the length of dictionary of letters/quantity available (key:value) | ||
| for letter, quantity in available_letters.items(): | ||
| # # randomly choose a letter in the dictionary using keys | ||
| # selection = random.choice(list(available_letters.keys()) -I was unable to get this to work in the loop- | ||
| selection = random.choice(letter) | ||
| # if the user list has < 10 letters | ||
| if len(hand_available) < 10: | ||
| # # if the quantity of the letter is not 0 (value) | ||
| if quantity != 0: | ||
| # # add the letter to the empty list | ||
| hand_available.append(selection) | ||
| # # subtract that number from the amounts available | ||
| quantity -= 1 |
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I like the idea here, but it unfortunately doesn't quite work. Here we're looking at the letters one at a time in the order they appeared in the dictionary. Because we're just looking at one letter at a time, random.choice doesn't actually choose anything randomly - it will just choose the one letter that we're looking at.
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| def uses_available_letters(word, letter_bank): | ||
| pass | ||
| for letter in word: | ||
| if letter in letter_bank: | ||
| return True | ||
| elif letter not in letter_bank or len(word) > len(letter_bank): | ||
| return False |
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This is on the right track! There are two main issues:
- It can return too early. If the first letter is in the letter bank, it will immediately return
True, without considering whether any of the other letters match. This is because the return statement on line 51 immediately ends our function, not letting the rest of the loop run. - It doesn't consider duplicate letters. If our word has two
As, but the letter bank only allows one, line 52 will not properly catch that we've used a letter too many times.
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This is my individual project as requested by Auberon. I was unable to collaborate in writing code with my partner so I started my own to practice.